674 Correction for Inertia.
curve AXB. Supposing the crank to revolve uniformly at eighty-
eight revolutions per minute, the velocity of crank pin,
v = 16-i ft. per sec.
that is, XY should measure 16-1. Dividing this ordinate into
16*1 parts will give the scale of piston velocity.
Next find the acceleration curve, QTR, adopting the method
already explained at p. 492, and illustrated in Fig. 454. QT will
shew, from base AB, the rate of increase of velocity, and TR the
rate of decrease, for the top diagram, viz., when the crank moves
through JKL; but on the return stroke, from B to A, lower
diagram, RT will be acceleration and TQ retardation. The
acceleration scale will not' be the same as the velocity scale,
but must be compressed in the ratio - as explained on p. 492.
In other words:
Reading on acceleration scale
= reading on velocity scale x —,
Produce x horizontally to Q. Then
_ ,. i6'i x 16-1 0
Reading A Q = ------;-------- = 148
By dividing AQ into 148 parts, an acceleration scale is there-
ore formed. (Seepp. 932 and 1107.)
Now the force required to produce a given acceleration in a
given mass (p. 473) is -f\ that is, the inertia force is propor-
tional to the acceleration. The weight of moving parts in this
engine is 8030 Ibs., and the inertia force at any moment,
_ wf 8030 . ,.
F = —- = —— x acceleration reading.
The acceleration curve may then be transformed into a curve
of inertia pressure (total) by multiplying by the above fraction or
by 8030 ~ 32*2 = 249'4, that is, the distance AQ must be divided
into 148 x 249*4 = 36,911 parts. This has been done along BP.
From the total pressure scale take 8030 Ibs., with dividers,
and move the curve QR down by that amount, to *f?, thus repre-
"eating the dead weight of the reciprocating parts.